Volume 25, Number 2 (May 2018), Pages 161–170
HYERS-ULAM STABILITY OF AN ADDITIVE
(ρ1, ρ2)-FUNCTIONAL INEQUALITY IN BANACH SPACES
Choonkil Park
aand Sungsik Yun
b,∗Abstract. In this paper, we introduce and solve the following additive (ρ1, ρ2)- functional inequality
kf(x+y+z)−f(x)−f(y)−f(z)k ≤ kρ1(f(x+z)−f(x)−f(z))k (0.1)
+kρ2(f(y+z)−f(y)−f(z))k, whereρ1 andρ2are fixed nonzero complex numbers with|ρ1|+|ρ2|<2.
Using the fixed point method and the direct method, we prove the Hyers- Ulam stability of the additive (ρ1, ρ2)-functional inequality (0.1) in complex Banach spaces.
1. Introduction and Preliminaries
The stability problem of functional equations originated from a question of Ulam [18] concerning the stability of group homomorphisms.
The functional equationf(x+y) =f(x) +f(y) is called theCauchy equation. In particular, every solution of the Cauchy equation is said to be anadditive mapping.
Hyers [10] gave a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers’ Theorem was generalized by Aoki [1] for additive mappings and by Rassias [16] for linear mappings by considering an unbounded Cauchy difference. A generalization of the Rassias theorem was obtained by G˘avruta [9] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias’
approach. The stability of quadratic functional equation was proved by Skof [17]
for mappingsf :E1 →E2, where E1 is a normed space andE2 is a Banach space.
Cholewa [7] noticed that the theorem of Skof is still true if the relevant domainE1 is replaced by an Abelian group.
Received by the editors January 04, 2018. Accepted May 04, 2018.
2010Mathematics Subject Classification. Primary 39B62, 47H10, 39B52.
Key words and phrases. Hyers-Ulam stability, additive (ρ1, ρ2)-functional inequality, fixed point method, direct method, Banach space.
∗Corresponding author.
°c2018 Korean Soc. Math. Educ.
161
Park [13, 14] defined additive ρ-functional inequalities and proved the Hyers- Ulam stability of the additive ρ-functional inequalities in Banach spaces and non- Archimedean Banach spaces. The stability problems of various functional equations have been extensively investigated by a number of authors (see [2, 6, 19]).
We recall a fundamental result in fixed point theory.
Theorem 1.1 ([3, 8]). Let (X, d) be a complete generalized metric space and let J :X → X be a strictly contractive mapping with Lipschitz constant α <1. Then for each given element x∈X, either
d(Jnx, Jn+1x) =∞
for all nonnegative integersn or there exists a positive integer n0 such that (1) d(Jnx, Jn+1x)<∞, ∀n≥n0;
(2) the sequence {Jnx} converges to a fixed point y∗ of J;
(3) y∗ is the unique fixed point of J in the setY ={y∈X |d(Jn0x, y)<∞};
(4) d(y, y∗)≤ 1−α1 d(y, Jy) for ally∈Y.
In 1996, G. Isac and Th.M. Rassias [11] were the first to provide applications of stability theory of functional equations for the proof of new fixed point theorems with applications. By using fixed point methods, the stability problems of several functional equations have been extensively investigated by a number of authors (see [4, 5, 15]).
In Section 2, we solve the additive (ρ1, ρ2)-functional inequality (0.1) and prove the Hyers-Ulam stability of the additive (ρ1, ρ2)-functional inequality (0.1) in Banach spaces by using the fixed point method.
In Section 3, we prove the Hyers-Ulam stability of the additive (ρ1, ρ2)-functional inequality (0.1) in Banach spaces by using the direct method.
Throughout this paper, letX be a real or complex normed space with normk · k and Y a complex Banach space with norm k · k. Assume that ρ1 and ρ2 are fixed nonzero complex numbers with|ρ1|+|ρ2|<2.
2. Additive (ρ
1, ρ
2)-functional Inequality (0.1):
a Fixed Point Method
In this section, we solve and investigate the additive (ρ1, ρ2)-functional inequality (0.1) in complex Banach spaces.
Lemma 2.1. If a mapping f :X →Y satisfies f(0) = 0and
kf(x+y+z)−f(x)−f(y)−f(z)k ≤ kρ1(f(x+z)−f(x)−f(z))k (2.1)
+kρ2(f(y+z)−f(y)−f(z))k for allx, y, z ∈X, then f :X→Y is additive.
Proof. Assume thatf :X→Y satisfies (2.1).
Letting x = y = z = 0 in (2.1), we get 2kf(0)k ≤ (|ρ1|+|ρ2|)kf(0)k and so f(0) = 0, since|ρ1|+|ρ2|<2.
Letting z= 0 in (2.1), we get
kf(x+y)−f(x)−f(y)k ≤0
for allx, y∈X. So f(x+y) =f(x) +f(y) for allx, y∈X. Thusf is additive. ¤ Using the fixed point method, we prove the Hyers-Ulam stability of the additive (ρ1, ρ2)-functional inequality (2.1) in complex Banach spaces.
Theorem 2.2. Let ϕ:X3 → [0,∞) be a function such that there exists an L < 1 with
ϕ
³x 2,y
2,z 2
´
≤ L
2ϕ(x, y, z) (2.2)
for allx, y, z ∈X. Let f :X →Y be a mapping satisfying f(0) = 0 and kf(x+y+z)−f(x)−f(y)−f(z)k ≤ kρ1(f(x+z)−f(x)−f(z))k (2.3)
+kρ2(f(y+z)−f(y)−f(z))k+ϕ(x, y, z) for all x, y, z ∈ X. Then there exists a unique additive mapping A : X → Y such that
kf(x)−A(x)k ≤ L
2(1−L)ϕ(x, x,0) for allx∈X.
Proof. Letting y=x and z= 0 in (2.3), we get kf(2x)−2f(x)k ≤ϕ(x, x,0) (2.4)
for all x∈X.
Consider the set
S :={h:X→Y, h(0) = 0}
and introduce the generalized metric onS:
d(g, h) = inf{µ∈R+:kg(x)−h(x)k ≤µϕ(x, x,0), ∀x∈X},
where, as usual, infφ= +∞. It is easy to show that (S, d) is complete (see [12]).
Now we consider the linear mappingJ :S →S such that Jg(x) := 2g
³x 2
´ for all x∈X.
Letg, h∈S be given such thatd(g, h) =ε. Then kg(x)−h(x)k ≤εϕ(x, x,0) for all x∈X. Hence
kJg(x)−Jh(x)k=
°°
°2g
³x 2
´
−2h
³x 2
´°°
°≤2εϕ
³x 2,x
2,0
´
≤2εL
2ϕ(x, x,0) =Lεϕ(x, x,0)
for all x∈X. Sod(g, h) =ε implies thatd(Jg, Jh)≤Lε. This means that d(Jg, Jh)≤Ld(g, h)
for all g, h∈S.
It follows from (2.4) that
°°
°f(x)−2f
³x 2
´°°
°≤ϕ
³x 2,x
2,0
´
≤ L
2ϕ(x, x,0) for all x∈X So d(f, Jf)≤ L2.
By Theorem 1.1, there exists a mapping A:X→Y satisfying the following:
(1)A is a fixed point ofJ, i.e.,
A(x) = 2A
³x 2
´ (2.5)
for all x∈X. The mapping Ais a unique fixed point ofJ in the set M ={g∈S:d(f, g)<∞}.
This implies that A is a unique mapping satisfying (2.5) such that there exists a µ∈(0,∞) satisfying
kf(x)−A(x)k ≤ µϕ(x, x,0) for all x∈X;
(2)d(Jlf, A)→0 as l→ ∞. This implies the equality
l→∞lim 2nf
³x 2n
´
=A(x)
for all x∈X;
(3)d(f, A)≤ 1−L1 d(f, Jf), which implies kf(x)−A(x)k ≤ L
2(1−L)ϕ(x, x,0) for all x∈X.
It follows from (2.2) and (2.3) that kA(x+y+z)−A(x)−A(y)−A(z)k
= lim
n→∞2n
°°
°°f
µx+y+z 2n
¶
−f
³x 2n
´
−f
³ y 2n
´
−f
³z 2n
´°°
°°
≤ lim
n→∞2n|ρ1|
°°
°°f
µx+z 2n
¶
−f
³x 2n
´
−f
³ z 2n
´°°°
° + lim
n→∞2n|ρ2|
°°
°°f
µy+z 2n
¶
−f
³ y 2n
´
−f
³ z 2n
´°°
°°+ lim
n→∞2nϕ
³x 2n, y
2n, z 2n
´
=kρ1(A(x+z)−A(x)−A(z))k+kρ2(A(y+z)−A(y)−A(z))k for all x, y, z∈X. So
kA(x+y+z)−A(x)−A(y)−A(z)k ≤ kρ1(A(x+z)−A(x)−A(z))k +kρ2(A(y+z)−A(y)−A(z))k for all x, y, z∈X. By Lemma 2.1, the mapping A:X →Y is additive. ¤ Corollary 2.3. Let r >1 and θ be nonnegative real numbers, and let f :X → Y be a mapping satisfying f(0) = 0and
kf(x+y+z)−f(x)−f(y)−f(z)k ≤ kρ1(f(x+z)−f(x)−f(z))k (2.6)
+kρ2(f(y+z)−f(y)−f(z))k+θ(kxkr+kykr+kzkr) for all x, y, z ∈ X. Then there exists a unique additive mapping A : X → Y such that
kf(x)−A(x)k ≤ 2θ 2r−2kxkr for allx∈X.
Proof. The proof follows from Theorem 2.2 by taking ϕ(x, y, z) = θ(kxkr+kykr+ kzkr) for all x, y, z∈X. ChoosingL= 21−r, we obtain the desired result. ¤ Theorem 2.4. Let ϕ:X3 → [0,∞) be a function such that there exists an L < 1
with
ϕ(x, y, z)≤2Lϕ
³x 2,y
2,z 2
´
for allx, y, z∈X. Letf :X→Y be a mapping satisfying f(0) = 0and(2.3). Then there exists a unique additive mapping A:X→Y such that
kf(x)−A(x)k ≤ 1
2(1−L)ϕ(x, x,0) for allx∈X.
Proof. Let (S, d) be the generalized metric space defined in the proof of Theorem 2.2.
Now we consider the linear mappingJ :S →S such that Jg(x) := 1
2g(2x) for all x∈X.
It follows from (2.4) that
°°
°°f(x)−1 2f(2x)
°°
°°≤ 1
2ϕ(x, x,0) for all x∈X.
The rest of the proof is similar to the proof of Theorem 2.2. ¤ Corollary 2.5. Let r <1 and θ be positive real numbers, and let f :X → Y be a mapping satisfyingf(0) = 0 and(2.6). Then there exists a unique additive mapping A:X→Y such that
kf(x)−A(x)k ≤ 2θ 2−2rkxkr for allx∈X.
Proof. The proof follows from Theorem 2.4 by taking ϕ(x, y, z) = θ(kxkr+kykr+ kzkr) for all x, y, z∈X. ChoosingL= 2r−1, we obtain the desired result. ¤ Remark 2.6. Ifρ is a real number such that|ρ1|+|ρ2|<2 andY is a real Banach space, then all the assertions in this section remain valid.
3. Additive (ρ
1, ρ
2)-functional Inequality (0.1): a Direct Method
In this section, we prove the Hyers-Ulam stability of the additive (ρ1, ρ2)-functional inequality (2.1) in complex Banach spaces by using the direct method.Theorem 3.1. Let ϕ:X3 →[0,∞) be a function such that Ψ(x, y, z) :=
X∞ j=1
2jϕ
³x 2j, y
2j, z 2j
´
<∞ (3.1)
for allx, y, z∈X. Letf :X→Y be a mapping satisfying f(0) = 0and(2.3). Then there exists a unique additive mapping A:X→Y such that
kf(x)−A(x)k ≤ 1
2Ψ(x, x,0) (3.2)
for allx∈X.
Proof. Letting y=x and z= 0 in (2.3), we get kf(2x)−2f(x)k ≤ϕ(x, x,0) (3.3)
and so
°°
°f(x)−2f
³x 2
´°°
°≤ϕ
³x 2,x
2,0
´
for all x∈X. Thus
°°
°2lf
³x 2l
´
−2mf
³ x 2m
´°°°≤
m−1X
j=l
°°
°2jf
³x 2j
´
−2j+1f
³ x 2j+1
´°°° (3.4)
≤
m−1X
j=l
2jϕ
³ x 2j+1, x
2j+1,0
´
for all nonnegative integersmand lwithm > l and allx∈X. It follows from (3.4) that the sequence {2kf(2xk)} is Cauchy for all x ∈X. Since Y is a Banach space, the sequence{2kf(2xk)} converges. So one can define the mappingA:X→Y by
A(x) := lim
k→∞2kf
³x 2k
´
for allx∈X. Moreover, letting l= 0 and passing the limitm→ ∞in (3.4), we get (3.2).
It follows from (2.3) and (3.1) that
kA(x+y+z)−A(x)−A(y)−A(z)k
= lim
n→∞2n
°°
°°f
µx+y+z 2n
¶
−f
³ x 2n
´
−f
³y 2n
´
−f
³ z 2n
´°°°
°
≤ lim
n→∞2n|ρ1|
°°
°°f
µx+z 2n
¶
−f
³x 2n
´
−f
³ z 2n
´°°
°° + lim
n→∞2n|ρ2|
°°
°°f
µy+z 2n
¶
−f
³ y 2n
´
−f
³ z 2n
´°°
°°+ lim
n→∞2nϕ
³x 2n, y
2n, z 2n
´
=kρ1(A(x+z)−A(x)−A(z))k+kρ2(A(y+z)−A(y)−A(z))k for all x, y, z∈X. So
kA(x+y+z)−A(x)−A(y)−A(z)k ≤ kρ1(A(x+z)−A(x)−A(z))k +kρ2(A(y+z)−A(y)−A(z))k for all x, y, z∈X. By Lemma 2.1, the mapping A:X →Y is additive.
Now, letT :X→Y be another additive mapping satisfying (3.2). Then we have kA(x)−T(x)k=
°°
°2qA
³x 2q
´
−2qT
³x 2q
´°°
°
≤
°°
°2qA
³x 2q
´
−2qf
³x 2q
´°°
°+
°°
°2qT
³x 2q
´
−2qf
³x 2q
´°°
°
≤2qΨ
³x 2q, x
2q,0
´ ,
which tends to zero asq → ∞ for all x∈X. So we can conclude that A(x) =T(x)
for all x∈X. This proves the uniqueness ofA. ¤
Corollary 3.2. Let r >1 and θ be nonnegative real numbers, and let f :X → Y be a mapping satisfying f(0) = 0 and (2.6). Then there exists a unique additive mapping A:X→Y such that
kf(x)−A(x)k ≤ 2θ 2r−2kxkr for allx∈X.
Theorem 3.3. Let ϕ:X3→[0,∞) be a function and let f :X →Y be a mapping satisfying f(0) = 0, (2.3) and
Ψ(x, y, z) :=
X∞ j=0
1
2jϕ(2jx,2jy,2jz)<∞
for all x, y, z ∈ X. Then there exists a unique additive mapping A : X → Y such that
kf(x)−A(x)k ≤ 1
2Ψ(x, x, ,0) (3.5)
for allx∈X.
Proof. It follows from (3.3) that
°°
°°f(x)−1 2f(2x)
°°
°°≤ 1
2ϕ(x, x,0) for all x∈X. Hence
°°
°°1
2lf(2lx)− 1
2mf(2mx)
°°
°°≤
m−1X
j=l
°°
°°1 2jf¡
2jx¢
− 1 2j+1f¡
2j+1x¢°
°°
°
≤
m−1X
j=l
1
2j+1ϕ(2jx,2jx,0) (3.6)
for all nonnegative integers m and l with m > l and all x ∈ X. It follows from (3.6) that the sequence {21nf(2nx)} is a Cauchy sequence for all x ∈ X. Since Y is complete, the sequence {21nf(2nx)} converges. So one can define the mapping A:X→Y by
A(x) := lim
n→∞
1
2nf(2nx)
for allx∈X. Moreover, letting l= 0 and passing the limitm→ ∞in (3.6), we get (3.5).
The rest of the proof is similar to the proof of Theorem 3.1. ¤ Corollary 3.4. Let r <1 and θ be positive real numbers, and let f :X → Y be a mapping satisfyingf(0) = 0 and(2.6). Then there exists a unique additive mapping A:X→Y such that
kf(x)−A(x)k ≤ 2θ 2−2rkxkr for allx∈X.
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aResearch Institute for Natural Sciences, Hanyang University, Seoul 04763, Republic of Korea
Email address: [email protected]
bDepartment of Financial Mathematics, Hanshin University, Gyeonggi-do 18101, Re- public of Korea
Email address: [email protected]